Solve the following simultanious equations: zy=28 and 2z-3y=13

zy=28 so y=28/z13=2z-3y13= 2z - (28 x 3)/z13=2z-84/zmultiply each side by z to give 13z = 2z2-84rearange for a quadratic2z2-13z-84=0solve by factorising(2z+8)(z-21/2)z= -4 0r 21/2substitute -4 into both initial equations-4y=28 and -8-3y=13 both give y=-7substitute 21/2 into both inital equations21y/2=28 and 21-3y=13 gives y=8/3(-4,-7) and (21/2,8/3)

OB
Answered by Oliver B. Further Mathematics tutor

1962 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

If the equation of a curve is x^2 + 9x + 8 = y, then differentiate it.


Make y the subject of the formula x = SQRT((y+1)/(y-2))


How would I solve the following equation d^2x/dt^2 + 5dx/dt + 6x = 0


Work out the gradient of the curve y=x^3(x-3) at the point (3,17)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences